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The boiling point of 0.1 mK(4)[Fe(CN)6] ...

The boiling point of `0.1 mK_(4)[Fe(CN)_6]` is expected to be `(K_(b)` for water = `0.52K kg mol^(-1)`)

A

`100.52^(@)C`

B

`100.10 4^(@)C`

C

`100.26^(@)C`

D

`102.6^(@)C`

Text Solution

Verified by Experts

The correct Answer is:
C
`Delta T_(b)=ixxk_(b)xxm`
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NARAYNA-SOLUTIONS & COLLIGATIVE PROPERTIES -EXERCISE : 2 (C.W)
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